A set containing a single non-zero vector is always a linearly independent subset! If av= 0 and v is not 0 then you must have a= 0. Do you see how that verifies the definition of "linearly independent"? Here, all you need to do is show that (0, 1, 1, 1, 0) is in V. becomes 0- 2(1)+ 3(1)- 1+ 2(0)= 0. Is that true?

(b) Extend S to a basis for V.

" " is a single linear equation in 5 unknown values. You can solve for any one of them, say, , leaving you "4 degrees of freedom"- that is your subspace has dimension 4 and a basis will require 4 vectors. Since you are already given one, you need to find 3 more that form an independent set.